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The $KE$ of an electron with de-Broglie wavelength of $0.3\: nm$ is

$\begin {array} {1 1} (a)\;0.168\: eV & \quad (b)\;16.8\: eV \\ (c)\;1.68\: eV & \quad (d)\;2.5\: eV \end {array}$

 

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And (b) 16.8 eV
$ \lambda=h/\sqrt2mE$
$\Rightarrow E = h^2 /2m\lambda^2$
$= (6.6 \times 10^{-34})^2 / [2 \times 9.1 \times 10^{-31} ) \times (0.3 \times 10^{-9} )^2 ]$
$= 2.65 \times 10^{-18} J = 16.8 eV$

 

answered Dec 24, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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